Question 922568
<pre>
You need to learn some trigonometric identities:

1.  sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
2.  sin(A-B) = sin(A)cos(B)-cos(A)sin(B) 
3.  cos(A+B) = cos(A)cos(B)-sin(A)sin(B)
4.  cos(A-B) = cos(A)cos(B)+sin(A)sin(B)
5.  {{{sec(A)=1/cos(A)}}}
5.  {{{csc(A)=1/sin(A)}}}
6.  {{{tan(A)=sin(A)/cos(A)}}}
7.  {{{cot(A)=cos(A)/sin(A)}}}
8.  {{{pi="180°"}}}
9.  {{{sin(n*pi) = 0}}} for all integers n
10. {{{cos(n*pi) = 1}}} for even integers n 
    {{{cos(n*pi) = -1}}} for odd integers n  
</pre> 
Suppose csc t = 4 and tan t > 0 
<pre>
4 is positive so csc is positive and tangent is positive,
so t is in quadrant 1.

The cosecant is {{{HYPOTENUSE/OPPOSITE}}}, so draw a right
triangle containing angle t.  Make the hypotenuse 4 and the 
side opposite angle t be 1.  Then csc(t) will be 4/1 or 4.

{{{drawing(400,8000/49,-.5,4.4,-.5,1.5,triangle(0,0,3.872983346,0,3.872983346,1), locate(2,0,ADJACENT), locate(.8,.2,t), locate(2,.75,4), locate(3.9,.6,1) 

)}}}

Since the hypotenuse = 4 and the opposite side = 1, then 

{{{HYPOTENUSE^2=OPPOSITE^2+ADJACENT^2}}}
{{{4^2=1^2+ADJACENT^2}}}
{{{16=1+ADJACENT^2}}}
{{{15=ADJACENT^2}}}
{{{sqrt(15)=ADJACENT}}}

So we put {{{sqrt(15)}}} on the adjacent side:

{{{drawing(400,8000/49,-.5,4.4,-.5,1.5,triangle(0,0,3.872983346,0,3.872983346,1), locate(2,-.1,sqrt(15)), locate(.8,.2,t), locate(2,.75,4), locate(3.9,.6,1) 

)}}}

</pre>
sin(t - 6&#960;) =
<pre>
First use 2 above.
{{{sin(t-6pi) = sin(t)cos(6pi)-cos(t)sin(6pi)}}} 
{{{sin(t-6pi) = (1/4)cos(6pi)-(sqrt(15)/4)sin(6pi)}}}
Now use 9 and 10 above 
{{{sin(t-6pi) = (1/4)(1)-(sqrt(15)/4)(0)}}}
{{{sin(t-6pi) = 1/4-0}}}
{{{sin(t-6pi) = 1/4}}}
</pre>
cot(5&#960;-t) = 
<pre>
First use 7 above
{{{cot(A)=cos(A)/sin(A)}}} 
{{{cot(5pi-t)=cos(5pi-t)/sin(5pi-t)}}}
Now use 4 and 2
{{{cot(5pi-t)=(cos(5pi)cos(t)+sin(5pi)sin(t))/(sin(5pi)cos(t)-cos(5pi)sin(t))}}}
Now use 9 and 10 and the right triangle to substitute for all the sines
and cosines
{{{cot(5pi-t)=((-1)(sqrt(15)/4)+0*(1/4))/(0*(sqrt(15)/4)-(-1)(1/4))}}}
{{{cot(5pi-t)=(-sqrt(15)/4)/(1/4)=-sqrt(15) }}}
</pre>
cos(-t) = 
<pre>
Write -t as 0-t and use 4

{{{cos(-t)=cos(0-t) = cos(0)cos(t)+sin(0)sin(t)}}}

Now use 9 and 10 and the right triangle to substitute for all the sines
and cosines

{{{cos(-t)=cos(0-t) = 1*sqrt(15)/4+(0)(1/4)}}}

{{{cos(-t)=sqrt(15)/4}}}
</pre> 
sin(2t) = 
<pre>
Write 2t as t+t and use 1:

{{{sin(2t)=sin(t+t) = sin(t)cos(t)+cos(t)sin(t)=2sin(t)cos(t)}}}

Use the right triangle above to substitute for the sines and cosines:

{{{sin(2t)=2(1/4)(sqrt(15)/4)}}}

{{{sin(2t)=(1/2)(sqrt(15)/4)}}}

{{{sin(2t)=(sqrt(15)/8)}}}
</pre>
cos(t+5&#960;/3) =
<pre>
Use 8

{{{cos(t+5pi/3)=cos(t+5("180°"/3))=cos(t+"300°")}}}

Use 3

{{{cos(t+5pi/3)=cos(t+"300°")= cos(t)cos("300°")-sin(t)sin("300°")}}}

300° is one of the special angles in quadrant IV with reference angle 60°

{{{cos("300°")=1/2}}}, {{{sin("300°")=-sqrt(3)/2}}}

{{{cos(t+5pi/3)= cos(t)cos("300°")-sin(t)sin("300°")}}

{{{cos(t+5pi/3)= (sqrt(15)/4)(1/2)-(1/4)(-sqrt(3)/2)}}}

{{{cos(t+5pi/3)=sqrt(15)/8+sqrt(3)/8=(sqrt(15)+sqrt(3))/8}}}

Edwin</pre>